216 ANALYTICAL MECHANICS Therefore the potential due to the entire sphere is V = F, 4- Vrt V H 3 a2 - ff2 2 a3 = — ym When .ft is plotted as abscissa and 7 as ordinate the distribution of the potential is given by a curve similar to (I) of Fig. 109. Now consider the intensity at a point in the field of the sphere. (a) POINT OUTSIDE THE SPHEEE. FIG. 109. dR m Therefore the distribution of the field intensity outside of the sphere is the same as that due to a particle placed at the center. (b) POINT WITHIN THE SPHERE. - dV m r> = -7-R. Therefore within the sphere the distribution of the field intensity obeys the harmonic law; i.e., the intensity varies directly as the distance from the center. In Fig. 109, curve (II) gives the distribution of the intensity of the field. PROBLEMS. 1. Find the potential and the field intensity due to a hollow sphere at a point (1) outside, (2) within the hollow part, and (3) in the solid part of the sphere. 2. Find the potential and the field intensity due to a circular disk of negligible thickness at a point on its axis. 3. Find the potential and the field intensity due to a straight wire of length I and mass m at a point on the axis of the wire. The cross-section of the wire is negligible.